EQUILIBRIUM OF RIGID BODIES 57
Combining (2) and (3), we obtain
D = /xd. (XIV)
The distance D is called the coefficient of rolling friction. Equation (XIV) states, therefore, that the coefficient of the rolling friction equals the coefficient of the sliding friction times the distance of the point of contact from the line of action of the force which urges the body to roll.
60. Friction Couple. — It is evident from the above equations that a change in the value of d does not affect the values of N and F, consequently it does not change the value of /*. This is as it should be, since, according to the laws of sliding; friction, /z depends only upon the nature of the surfaces in contact. A change in d, however, changes the value of D;; in other words, it changes the point of application of R.. When d = 0, that is, when S is applied at the point of contact, D = 0, in which case the body is urged to slide only. But when d is not zero the force S not only urges the body to slide but also to roll; therefore, in addition to the resisting force F, a resisting torque comes into play. This torque, which is due to the couple formed by N and W, is called the friction couple.
PROBLEMS.
1. A gig is so constructed that when the shafts are horizontal the center of gravity of the gig is over the axle of the wheels. The gig rests on perfectly rough horizontal ground. Find the least force which, acting at the ends of the shafts, will just move the gig.
2. Find the smallest force which, acting tangentially at the rim of a flywheel, will rotate it. The weight and the radius of the flywheel, the radius of the shaft, and the coefficient of friction between the shaft and its bearings are supposed to be known.
3. A flywheel of 500 pounds weight is brought to the point of rotation by a weight of 10 pounds suspended by means of a string wound around its rim. Find the coefficient of friction between the axle and its bearings. The diameters of the wheel and the axle are 10 feet and 8 inches, respectively.